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[some formatting tweaked, and the question copied from the title to the main body, by YC]


Hi,

I've been struggling a lot to calculate this integral.

$$ \int_0^\infty \frac{k^{n-1}}{\prod_{i=1}^n (k^2+ x_i^2)}\; dk $$ where $x_i$ are constants and $n\geq 1$.

I did the calculation for n=1,2,3,4, with the hope of identifying some form and then find the result by induction. But here is what I got:

  • n=1: I= (pi/2) * abs(x1)

  • n=2: I= (1/2) * 1/(x2ˆ(2)-x1ˆ(2)) * log(x2ˆ(2) / x1ˆ(2))

  • n=3: I= (pi/2) * [abs(x1) (x2ˆ(2)-x3ˆ(2)) +abs(x2) (x3ˆ(2)-x1ˆ(2))+ abs(x3) (x1ˆ(2)-x2ˆ(2))] / [(x2ˆ(2)-x3ˆ(2) (x3ˆ(2)-x1ˆ(2)) (x1ˆ(2)-x2ˆ(2)]

  • n=4: I= (1/2) * [ A1 log(x1ˆ(2)) + A2 log(x2ˆ(2)) +... A4 log(x4ˆ(2))), where Ai= xiˆ(2) / [ prod (xjˆ(2)-xiˆ(2))]

-->> This makes me think that the result depends on whether n is even or uneven; that is, we would have a form in log( ) for n even, and something in pi/2 for n uneven?

Could you please help me here? What is the correct result and how to get it?

Your help is so much appreciated, many many thanks in advance! Elise

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closed as off-topic by Ricardo Andrade, Chris Godsil, Ryan Budney, Andrey Rekalo, Dmitri Pavlov Jan 20 at 16:19

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1 Answer 1

up vote 2 down vote accepted

You are correct. Use the partial fraction decomposition: http://en.wikipedia.org/wiki/Partial_fraction For example, if $n=4$, the decomposition is (over the rationals):

$$\begin{array}{l} {\frac {ck}{ \left( {k}^{2}+c \right) \left( -c+a \right) \left( -c+ b \right) \left( -d+c \right) }}-\\\ {\frac {dk}{ \left( {k}^{2}+d \right) \left( -d+a \right) \left( -d+b \right) \left( -d+c \right) }}-\\\ {\frac {bk}{ \left( {k}^{2}+b \right) \left( -b+a \right) \left( -c+b \right) \left( -d+b \right) }}+\\\ {\frac {ak}{ \left( {k}^{2}+a \right) \left( -b+a \right) \left( -c+a \right) \left( -d+a \right) }}\end{array} $$ where $a=x_1^2,b=x_2^2,...$. I guess you can get the pattern. For odd $n$ there is no $k$ in the numerator. This is the cause of the difference you noticed.

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Thank you so much for your insights, much appreciated :-) Best, Elise –  Payze Apr 2 '11 at 11:47

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